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Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Solution
A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is
A
r tan yo cm
B
r sin yo cm
C
r cot yo cm
D
r cos yo cm
E
r cosec yo cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{r}{h}\) = tan yo
h = \(\frac{r}{tan y^o}\)
= r cot yo
Source: Myschool Past Questions Archive (Record #32438)
Mathematics
UTME / JAMB 1984
Permutations, Combinations & Probability
Solution
Two fair dice are rolled. What is the probability that both show up the same number of points.
A
\(\frac{1}{36}\)
B
\(\frac{7}{36}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{3}\)
E
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #32439)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
The larger value of y for which (y - 1)2 = 4y - 7 is
A
2
B
4
C
6
D
7
E
8
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4
Source: Myschool Past Questions Archive (Record #32440)
Mathematics
UTME / JAMB 1984
Trigonometry, Elevation & Bearings
Solution
If sin \(\theta\) = \(\frac{x}{y}\) and 0o o then find \(\frac{1}{tan\theta}\)
A
\(\frac{x}{\sqrt{y^2 - x^2}}\)
B
\(\frac{y}{x}\)
C
\(\frac{\sqrt{y^2 + x^2}}{y^2 - x^2}\)
D
\(\frac{y^2 - x^2}{x}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\)
sin\(\theta\) = \(\frac{x}{y}\)
cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)
Source: Myschool Past Questions Archive (Record #32441)
Mathematics
UTME / JAMB 1984
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below
\(\begin{array}{c|c} \text{Class boundary in cm} & \text{frequency} & \\\hline 3.35 - 3.45 & 3\\ 3.45 - 3.55 & 6\\ 3.55 - 3.65 & 7\\ 3.65 - 3.75 & 4\end{array}\)
What is the mean diameter of the copper spheres?
A
3.40cm
B
3.58cm
C
3.56cm
D
3.62cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\begin{array}{c|c} \text{x(mid point)} & f & fx\\ \hline 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0\\3.6 & 7 & 25.2\\3.7 & 4 & 14.8\end{array}\)
\(\sum f\) = 20
\(\sum fx\) = 71.2
mean = \(\frac{\sum fx}{\sum f}\)
= \(\frac{71.2}{20}\)
= 3.56
Source: Myschool Past Questions Archive (Record #32442)
Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
What is the volume of this regular three dimensional figure?
Attached Figure / Diagram
A
160cm2
B
48cm2
C
120cm2
D
40cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Volume of the three dimensional figures = v = A x h
A = \(\frac{1}{2}\) x 4 x 3
= 6cm2
V = 6 x 8
= 48cm2
Source: Myschool Past Questions Archive (Record #32776)
Mathematics
UTME / JAMB 1984
Indices, Logarithms & Surds
Diagram
Solution
Find the area of the shaded portion of the semicircular figure.
Attached Figure / Diagram
A
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
B
\(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
C
\(\frac{1}{2}r^2 \pi\)
D
\(\frac{1}{8}r^2 \sqrt{3}\)
E
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Asector = \(\frac{60}{360} \times \pi r^2\)
= \(\frac{1}{6} \pi r^2\)
A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\)
\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)
A\(\text{shaded portion}\) = Asector -
A\(\bigtriangleup\)
= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)
= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)
= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
Source: Myschool Past Questions Archive (Record #32778)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS
Attached Figure / Diagram
A
60o
B
90o
C
120o
D
30o
E
80o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
From the diagram, PQRSTW is a regular hexagon.
Hexagon is a six sided polygon.
Sum of interior angles of polygon = (2n - 4)90o = [2 x 6 - 4] x 90 = 8 x 90 = 720o
each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)
Source: Myschool Past Questions Archive (Record #32779)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, QRS is a line, PSQ = 35o, SPR = 30o and O is the centre of the circle. Find OQP.
Attached Figure / Diagram
A
35o
B
30o
C
130o
D
25o
E
65o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
< PSQ = 35o, < SPR = 30o
O is the centre of the circle, PRQ = 65o
< OQP = 90 - 65
= 25o
Source: Myschool Past Questions Archive (Record #32780)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, determine the angle marked y
Attached Figure / Diagram
A
66o
B
110o
C
26o
D
70o
E
44o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
From the diagram, o, o
o - (70 + 44o) = 66o
Source: Myschool Past Questions Archive (Record #32781)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
Find the x co-ordinates of the points of intersection of the two equations in the graph
Attached Figure / Diagram
A
1, 1
B
0, 4
C
0,0
D
4, 0
E
4, 9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
If y = 2x + 1 and y = x2 - 2x + 1
then x2 - 2x + 1 = 2x + 1
x2 - 4x = 0
x(x - 4) = 0
x = 0 or 4
Source: Myschool Past Questions Archive (Record #32782)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP
Attached Figure / Diagram
A
4cm
B
16cm
C
9cm
D
14cm
E
Impossible, insufficient data
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{PQ}{6 + RT} = \frac{6}{12} = \frac{6}{PS}\)(Similar triangles)
\(\frac{6}{12} = \frac{8}{PS}\)
PS = \(\frac{12 \times 8}{6}\)
= 16cm
Source: Myschool Past Questions Archive (Record #32783)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, O is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is
Attached Figure / Diagram
A
67\(\frac{1}{2}\)o
B
22\(\frac{1}{2}\)o
C
33\(\frac{1}{2}\)o
D
45o
E
90o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
< R = 180o - 45o (sum of angles on a straight line)
< R = < P = 45o (corresponding angles)
< PSO = < P = 45o (\(\bigtriangleup\)PSO is a right angle)
Source: Myschool Past Questions Archive (Record #32784)
Mathematics
UTME / JAMB 1984
Indices, Logarithms & Surds
Diagram
Solution
XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.
Attached Figure / Diagram
A
5\(\sqrt{3}\)cm
B
3\(\sqrt{5}\)cm
C
3\(\sqrt{3}\)cm
D
5cm
E
6cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
by Pythagoras's theorem, XW = 3cm
Also by Pythagoras theorem, XY2 = 62 + 32
XY2 = 36 + 9 = 45
XY = \(\sqrt{45} = 3 \sqrt{5}\)
Source: Myschool Past Questions Archive (Record #32785)
Mathematics
UTME / JAMB 1983
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize completely 81a\(^4\) - 16b\(^4\)
A
(3a + 2b)(2a - 3b)(9a2 + 4b2)
B
(3a - 2b)(2a - 3b)(4a2 - 9b2)
C
(3a - 2b)(3a + 2b)(9a2 + 4b2)
D
(6a - 2b)(8a - 3b)(4a3 - 9b2)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
81a\(^4\) - 16b\(^4\) = (9a\(^2\))\(^2\) - (4b\(^2\))\(^2\)
= (9a\(^2\) + 4b\(^2\))(9a\(^2\) - 4b\(^2\))
N:B 9a\(^2\) - 4b\(^2\) = (3a - 2b)(3a + 2b)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32471)